Three-Phase TLINE - Time Domain

Electricity and Magnetism Level 4

A balanced three-phase voltage source feeds a balanced set of load resistors through a lossless transmission line. The three phases of the transmission line are magnetically coupled, which can be modeled in the time-domain using self and mutual inductances as follows:

$V_{AS} (t) - L_S \dot{I}_A - L_M \dot{I}_B - L_M \dot{I}_C = V_{AR}(t) \\ V_{BS}(t) - L_M \dot{I}_A - L_S \dot{I}_B - L_M \dot{I}_C = V_{BR}(t) \\ V_{CS}(t) - L_M \dot{I}_A - L_M \dot{I}_B - L_S \dot{I}_C = V_{CR}(t)$

In the circuit, the star (neutral) point of the voltage sources is taken as the voltage reference. By convention, the currents flow across the transmission line from left to right. The system parameters are:

$V_{AS}(t) = 100 \, \sin(\omega t) \\ V_{BS}(t) = 100 \, \sin(\omega t - 2 \pi /3) \\ V_{CS}(t) = 100 \, \sin(\omega t + 2 \pi /3) \\ \omega = 120 \pi \\ L_S = 50/\omega \\ L_M = 30/\omega \\ R = 1$

At time $t = 0$ , currents $(I_A, I_B, I_C)$ are all zero. Let $I_1$ be the maximum (greatest) instantaneous value of $I_A$ over all time. Let $I_2$ be the AC-steady-state peak value of $I_A$ , after all of the initial transients have died out.

What is $I_1 + I_2$ ?

Note: $I_1$ and $I_2$ are both positive numbers

The answer is 14.27.

1 solution

Karan Chatrath
Nov 14, 2019

The first step is writing down the circuit equations by applying Kirchoff's laws.

$V_{AS} - I_AR = (V_{AS} - V_{AR})$ $V_{BS} - I_BR = (V_{BS} - V_{BR})$ $V_{CS} - I_CR = (V_{CS} - V_{CR})$

Replacing the equation $V_{AS} - V_{AR}$ by the given transmission line model, in each equation, re-writing in a matrix form, and simplifying gives:

$\left[\begin{matrix} \dot{I}_{A} \\\dot{I}_{B} \\\dot{I}_{C} \end{matrix}\right] = -R \left[\begin{matrix} L_S&L_M&L_M\\L_M&L_S&L_M\\L_M&L_M&L_S \end{matrix}\right]^{-1} \left[\begin{matrix} I_{A} \\I_{B} \\I_{C} \end{matrix}\right] + \left[\begin{matrix} L_S&L_M&L_M\\L_M&L_S&L_M\\L_M&L_M&L_S \end{matrix}\right]^{-1}\left[\begin{matrix} V_{AS} \\V_{BS} \\V_{CS} \end{matrix}\right]$

In shorthand notation:

$\boxed{\dot{I} = AI + BV}$ $A = -R \left[\begin{matrix} L_S&L_M&L_M\\L_M&L_S&L_M\\L_M&L_M&L_S \end{matrix}\right]^{-1} \ ; \ B = \left[\begin{matrix} L_S&L_M&L_M\\L_M&L_S&L_M\\L_M&L_M&L_S \end{matrix}\right]^{-1}$ $I = \left[\begin{matrix} I_{A} \\I_{B} \\I_{C} \end{matrix}\right] \ ; \ V = \left[\begin{matrix} V_{AS} \\V_{BS} \\V_{CS} \end{matrix}\right]$ $I_A(0) = 0 \ ; \ I_B(0) = 0 \ ; \ I_C(0) = 0$

The given set of linear differential equations can be solved analytically, however, a numerical route is chosen. The coding details are left out for now but the solution $I_A(t)$ looks as such:

One can see that the transient response dies out after approximately 0.2 seconds. From this array of time-varying data as shown in the plot, the values asked for can be obtained.

The peak current is 9.2715 A while the peak steady-state current is 4.9946 A (almost 5 A). The required answer is $\boxed{14.2662 \ A}$ .

Three-Phase TLINE - Time Domain

The answer is 14.27.

1 solution

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